De Giorgi-Nash-Moser inequality

In parabolic PDEs, if we apply De Giorgi-Nash-Moser estimate, we usually encounter the following inequality \[ Q_k \leq \f{2^k}{M}Q_{k-1}^{1+\ep},\qquad \forall\,k\geq 1, \] for some $\ep, M>0$, and we then conclude from this inequality that $Q_k$ is uniformly bounded if $M$ is large enough, which is the goal of this note. **Lemma 1.** Fix $\ep>0$ and let $Q_k$ be a positive sequence such that \[ Q_k \leq \f{2^k}{M}Q_{k-1}^{1+\ep},\qquad \forall k\geq 1, \] for some $M>0$. Then if \[ M = Q_0^{\ep/(1+\ep)} 2^{1/\ep}, \] then \[ Q_k \leq \f{M^{1/\ep}}{2^{1/\ep^2}} = \f{Q_0}{ e^{1+\ep} }, \] and if \[ M> Q_0^{\ep/(1+\ep)} 2^{1/\ep}. \] then \[ Q_k \rightarrow 0,\qquad k\rightarrow +\infty. \] //Proof.// Let $q_k=\ln(Q_k)$ and we have \begin{equation} (1)\qquad q_k \leq (1+\ep)q_{k-1} + k\ln 2 - \ln M. \end{equation} Iterating $(1)$, we find \begin{equation} \begin{split} q_k&\leq (1+\ep)q_{k-1} + k\ln 2 - \ln M ~\\ &\leq (1+\ep)\big[ (1+\ep)q_{k-2} + (k-1)\ln 2 - \ln M \big]+ k\ln 2 - \ln M ~\\ &\leq (1+\ep)^2q_{k-2} + (k + (k-1)(1+\ep) )\ln 2 - (1 + (1+\ep))\ln M ~\\ &\leq \cdots ~\\ &\leq (1+\ep)^k q_0 +\big(\sum_{i=0}^k (k-i)(1+\ep)^i \big)\ln 2 - \big(\sum_{i=0}^k (1+\ep)^k\big)\ln M. \end{split} \end{equation} We directly compute \[ \sum_{i=0}^k (1+\ep)^k = \f{ (1+\ep)^{k+1} - 1}{\ep},\qquad \sum_{i=0}^k (k-i)(1+\ep)^i = \f{ (1+\ep)^{k+1} - 1}{\ep^2}-\f{k+1}{\ep}. \] Hence, we obtain \[ q_k\leq (1+\ep)^k q_0 + \f{ (1+\ep)^{k+1} - 1}{\ep}\big( \f{\ln 2}{\ep} - \ln M \big), \] which implies that \[ \begin{split} (2)\qquad Q_k&\leq Q_0^{ (1+\ep)^k } \big(\f{2^{1/\ep}}{M}\big)^{(1+\ep)^k(1+\ep)/\ep} \big(\f{2^{1/\ep}}{M}\big)^{-1/\ep}~\\ &\leq \f{M^{1/\ep}}{2^{1/\ep^2}} \bigg(\f{Q_0 2^{(1+\ep)/\ep^2} }{M^{(1+\ep)/\ep}}\bigg)^{(1+\ep)^k}. \end{split} \] Observing that \[ Q_0 2^{(1+\ep)/\ep^2} \leq M^{(1+\ep)/\ep} \qquad \Longleftrightarrow\qquad M\geq Q_0^{\ep/(1+\ep)} 2^{1/\ep}, \] we conclude from $(2)$ that the claims in **Lemma 1** hold. This ends the proof. $ \square $

Counter example: a function has second order weak derivative but does not have first order weak derivataive

Give a counter example of a function $u(x,y)$ so that the first order weak derivative of $u(x,y)$ does not exist, but the second order weak derivative $\partial_{xy} u$ exists. The general counter example is the following: Suppose that $f(x)$ is a continuous but not absolutely continuous function on $[0,1]$, then we know $f(x)$ does not have first order weak derivative (otherwise $f(x)$ is absolutely continuous). Then we set $u(x,y) = f(x) + f(y)$, and it is clear that $u(x,y)$ does not have first order weak derivative. However, we can show that the second order weak derivative $\partial_{xy}u$ exists and equals $0$. A typical continuous but not absolutely continuous function is the Cantor function. === References [[|Why is the cantor function not absolutely continuous]]

Hand-Written Notes

#A note about the derivation of the Navier-Stokes equations and it is written when I was taking the reading course with Prof. David Hoff in Indiana University. [[pdf/Derivation-NSE.pdf|Download]] #The following notes are aiming to understand the famous paper titled" Well-posedness for the Navier-Stokes equations" by H. Koch and D. Tataru. Download [[pdf/Koch-Tataru-NSE-1.pdf|Note 1]], [[pdf/Koch-Tataru-NSE-2.pdf|Note 2]], [[pdf/Koch-Tataru-NSE-2.pdf|Note 3]] and see also [[post/196|Leray projection - NSE Notes I]], [[post/197|Kernel estimates I - NSE notes II]], [[post/198|Kernel estimates II - NSE notes II]], [[post/199|A pointwise estimate - NSE notes V]], [[post/200|the invariance of BMO^{-1}-norm - NSE notes III]], [[post/201|Fixed point theorem - NSE notes IV]]